I thought it does not have absolute maximum, but wanted to just check and see why

  • 4
    $\begingroup$ it's a continuous function on a closed interval, so it better have a maximum $\endgroup$ – ShawnD Mar 13 '12 at 0:06
  • $\begingroup$ Your function has to have an absolute maximum, by Weierstrass theorem. $\endgroup$ – Pacciu Mar 13 '12 at 0:07
  • $\begingroup$ I think I might have got confused with the local maxima $\endgroup$ – Jeremy Carlos Mar 13 '12 at 0:13

A continuous function on a closed interval always has both an absolute max and an absolute min. And, the absolute max and min can only occur at critical points or the end points of the interval (I know of at least one book that includes end points as critical points). Therefore, the strategy that calculus books give is to find all critical points (in your interval) and plug these, and the end points, in to the original. The highest value you get is your absolute max and the smallest value you get is your absolute min (and you get the $x$ values where the absolute max and min occur, as the $x$ you plugged in). Since they must occur, and since these are the only possible places where they can occur, this guarantees we will find the absolute max and min if we do this.

If $f(x) = \sin^2 x - \sin x$, then $f'(x) = 2\sin x \cos x - \cos x = \cos x(2\sin x - 1)$. This is always defined so the only critical points will be where this is 0. A product is 0 exactly when one of the two parts is 0, so this is 0 when $\cos x = 0$ or when $\sin x = \frac{1}{2}$. In $\left[0, \frac{3\pi}{2} \right]$, the solutions would be: $\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3\pi}{2}$, $\sin x = \frac{1}{2}$ when $x = \frac{\pi}{6}$ and $x = 5\pi / 6$.

Now, plugging in all these values, and the end points gives:

$\begin{align*} f(0) &= 0 \\ f(\pi/6) &= - \frac{1}{4} \\ f(\pi/2) &= 0 \\ f(5\pi/6) &= - \frac{1}{4} \\ f(3\pi/2) &= 2 \end{align*}$

Therefore, the max is 2 at $3\pi/2$ and the min is $-\frac{1}{4}$ at both $\pi/6$ and $5\pi/6$.

  • $\begingroup$ One small point: when we find the critical numbers, we use only those that lie in the interval. $\endgroup$ – Patrick Mar 13 '12 at 3:32
  • $\begingroup$ @Patrick Yes, thanks, I didn't make that clear in my description. But, I did use that in my solution. $\endgroup$ – Graphth Mar 13 '12 at 12:00

We are interested in $w^2-w$, with the proviso that $-1\le w\le 1$. Completing the square, we find that $4w^2-4w=(2w-1)^2+3$. So the minimum value of $4w^2-4w$ is $3$, reached when $w=1/2$. The maximum value in our interval is reached when $w$ is as negative as possible, namely at $w=-1$.

So the minimum value of $w^2-w$ is $\frac{3}{4}$, reached when $\sin x=\frac{1}{2}$, that is, at $x=\frac{\pi}{6}$ and also at $x=\pi-\frac{\pi}{6}$.

The maximum value of $w^2-w$ is $\frac{12}{4}$, reached when $\sin x=-1$, that is, at $x=\frac{3\pi}{2}$.

Remark: You came close to being right on no absolute maximum. If we replace the interval $[0,\frac{3\pi}{2}]$ by $(0,\frac{3\pi}{2})$, there is no longer an absolute maximum.


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